Problem: Simplify the following expression: $y = \dfrac{-2x^2- 1x+28}{-2x + 7}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-2)}{(28)} &=& -56 \\ {a} + {b} &=& &=& {-1} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-56$ and add them together. Remember, since $-56$ is negative, one of the factors must be negative. The factors that add up to ${-1}$ will be your ${a}$ and ${b}$ When ${a}$ is ${7}$ and ${b}$ is ${-8}$ $ \begin{eqnarray} {ab} &=& ({7})({-8}) &=& -56 \\ {a} + {b} &=& {7} + {-8} &=& -1 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-2}x^2 +{7}x) + ({-8}x +{28}) $ Factor out the common factors: $ x(-2x + 7) + 4(-2x + 7)$ Now factor out $(-2x + 7)$ $ (-2x + 7)(x + 4)$ The original expression can therefore be written: $ \dfrac{(-2x + 7)(x + 4)}{-2x + 7}$ We are dividing by $-2x + 7$ , so $-2x + 7 \neq 0$ Therefore, $x \neq \frac{7}{2}$ This leaves us with $x + 4; x \neq \frac{7}{2}$.